4a. Use the Nernst equation to calculate the cell voltage (E) for the following redox reaction:
i. Fe3+(aq)+Cu(s)→Cu2+(aq)+Fe2+(aq), given that [Fe3+]=0.05 and [[Cu2+]=0.125M at 25°C, and
Fe3++e→Fe2+ E0=0.77 V
Cu2+2e→Cu E0=0.34 V
4b. Use the information provided in question 4a to calculate the change in free energy (ΔG) and change in entropy (ΔS) for the redox reaction:
i. Fe3+(aq)+Cu(s)→Cu2+(aq)+Fe2+(aq)
What do the ΔG and ΔS values indicate about the spontaneity of the redox reaction?
2Fe3+(aq)+Cu(s) ----> Cu2+(aq)+2Fe2+(aq),
E0cell = E0cathode - E0anode
= (0.77)-(0.337)
= 0.433 v
Ecell = E0cell -
(0.0591/n)log([Cu2+(aq)][Fe2+(aq)]^2/[Fe3+]^2)
= 0.433 - (0.0591/2)log(0.125* 0.05^2/1)
= 0.54 v
DG0 = - nFE0cell
= -2*96500*0.433
= -83.57 kj/mol
as DG0 = -ve, the process is spontaneous.
DG = - nFEcell
= -2*96500*0.54
=-104.22 kj/mol
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